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46916926 Testing Euclidean Geometry

Dec 14, 2015

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Dear Alfio,

THE TRISECTION OF ANY ANGLE This article has been written during Markos Georgallides : Tel-00357 -99 634628last month of 2010 and is solely for Civil Engineer(NATUA) : Fax-00357-24 653551

my own amusement and private study 38 , Z .Kitieos St , 6022 , Larnaca and also to all those readers that believe Expelled from Famagusta town occupied to Euclid Geometry as the model of by the Barbaric Turks . all nature . A meter also for testing sufficiency of Geometries . Email < [email protected] >1. Archimedes method 2. Pappus method Consider the angle < AOB .

Draw circle ( O , OA ) with its center at the vertex O and produce side BO to D .

Insert a straight line AD so point C is on the circle and point D on line BO and length DC such that it is equal to the radius of the circle . Proof : Since CD = CO then triangle CDO is isosceles and angle < CDO = COD The external angle OCA of triangle CDO is < OCA = CDO + COD = 2. CDO

and equal to angle ADO and since angle < OAC = OCA then < OAC = 2.ODA

The external angle AOB of triangle OAD is < AOB = OAD + ODA =

2. ODA + ODA = 3 . ODA 2. Pappus method :

It is a slightly different of Archimedes method can be reduced to a neusis as follows

Consider the angle < AOB .

Draw AB perpendicular to OB .

Complete rectangle ABOC .

Produce the side CA to E .

Insert a straight line ED of given length 2 .OA between AE and AB

in such a way that ED verges towards O . Then angle < AOB = 3 . DOB 3. The Present method : We extend Archimedes method as follows : a . ( F. 4 ) Given an angle < AOB = AOC = 90

1. Draw circle ( A , AO = OA ) with its center at the vertex A intersecting

circle ( O , OA = AO ) at the points A1 , A2 respectively .

2 . Produce line AA1 at C so that A1C = A1 A = AO and draw AD // OB .

3 . Draw CD perpendicular to AD and complete rectangle AOCD .

4 . Point F is such that OF = 2 . OA b . ( F. 5-6-7 ) Given an angle < AOB < 90 1. Draw AD parallel to OB .2. Draw circle ( A , AO = OA ) with its center at the vertex A intersecting circle

( O , OA = AO ) at the points A1 , A2 .

3. Produce line A A1 at D1 so that A1D1 = A1 A = OA . 4 . Point F is such that OF = 2 . OA = 2 . OAo .

5 . Draw CD perpendicular to AD and complete rectangle AOCD .6 . Draw Ao E Parallel to A C at point E ( or sliding E on OC ) .

7 . Draw AoE parallel to OB and complete rectangle AoOEE .

8 . Draw AF intersecting circle ( O , OA ) at point F1 and insert on AF

segment F1 F2 equal to OA F1 F2 = OA .

9 . Draw AE intersecting circle ( O , OA ) at point E1 and insert on AE

segment E1 E2 equal to OA E1 E2 = OA = F1 F2 . Show that :

a) For all angles equal to 90 Points C and E are at a constant distance OC = OA . 3 and OE = OAo . 3 , from vertices O , and also AC // AoE . b) The geometrical locus of points C , E is the perpendicular CD , EE .

c) All equal circles with their center at the vertices O , A and radius OA = AO have the same geometrical locus EE OE for all points A on AD , or All radius of equal circles drawn at the points of intersection with its

Centers at the vertices O , A and radius OA = AO lie on CD , EE . d) Angle < D1OA is always equal to 90 and angle AOB is created by rotation

of the right-angled triangle AOD1 through vertex O . e ) Angle < AOB is created in two ways , By constructing circle ( O , OA = OAo ) and by sliding of point A on line A D Parallel to OB from point A to A . f) The rotation of lines AE , AF on circle ( O , OA = OAo ) from point E to point F which lines intersect circle ( O , OA ) at the points E1 , F1

respectively , fixes a point G on line EF and a point G1 common to line AG and to the circle ( O , OA ) such that G G1 = OA .

Proof : a ) .. ( F3 , F4 ) Let OA be one-dimentional Unit perpendicular to OB such that angle

< AOB = AOC = 90Draw the equal circles ( O,OA ) , ( A , AO ) and let points A1 , A2 be the points of intersection . Produce AA1 to C . Since triangle AOA1 has all sides equal to OA ( AA1 = AO = OA1 ) then it is an equilateral triangle and angle < A1AO = 60 Since Angle < CAO = 60 and AC = 2. OA then triangle ACO is right-angled and angle < AOC = 90 , so the angle ACO = 30 Complete rectangle AOCD

Angle < ADO = 180 90 60 = 30 = ACO = 90 / 3 = 30

From Pythagoras theorem AC = AO + OC or OC = 4.OA - OA = 3. OA

and OC = OA . 3 . For OA = OAo then AoE = 2. OAo and OE = OAo . 3 . Since OC / OE = OA / OAo then line CA is parallel to EAob ) .. ( F5 , F6 )

Triangle OAA1 is isosceles , therefore angle < A1AO = 60 . Since A1A = A1D 1 and angle < A1AO = 60 then triangle AOD 1 is right-angle triangle and angles < D 1OA = 90 , angle < OD 1A = 30 . Since the circle of diameter D1A passes through point O and also through the foot of the perpendicular from point D1 to AD , and since also ODA = ODA = 30 then this foot point coincides with point D , therefore the locus of point E is the perpendicular EE on OC . For AA1 > A1D 1 is easy to see that D1 is on the perpendicular D1 E on OC . c ) .. ( F5 , F6 )Since the Parallel from point A 1 to OA passes through the middle of OD 1 , and in case where AOB = AOC = 90 through the middle of AD then the circle with diameter D1A passes through point D which is the base of the perpendicular , i.e.The geometrical locus of points C , E is the perpendicular CD , EE . d ) .. ( F5 , F6 ) Since A1A = A1D 1 and angle < A1AO = 60 then triangle AOD 1 is right-angle triangle and angle < D 1OA = 90 .Since angle < AD 1O is always equal to 30 and angle D1OA is always equal to 90 so angle AOB is created by the rotation of right - angled triangle AOD1 through vertex O .

Since tangent through Ao to circle ( O , OA) lies on the circle of half radius OA then this is perpendicular to OA and equal to AA .

F2F1 = OA A1E = OAo GA1 = OA E2E1 = OA e) .. ( F5 , F6 , F.7a ) Let point G be sliding on OB between points E and F where lines AE , AG , AF

intersect circle ( O , OA ) at the points E1 , G 1 , F1 respectively and then exists

FF1 > OA , GG1 = OA , E E1 < OA . Points E , F are limiting points of rotation of lines AE , AF ( because for angle

< AOB = 90 A1C = OA , A1Ao = A1E and for angle < AOB = 0 OF = 2 . OA ) and also E1E2 = OA , F1F2 = OA and point G1 common to circle ( O , OA ) and line AG such that GG1 = OA .

When point G1 is moving ( rotated ) on circle ( E2 , E2E1 = OA ) then GG1 must

be stretched to G ( where then G is not on line OB ) so that GG1 = OA .

When point G1 is moving ( rotated ) on circle ( F2 , F2F1 = OA ) then GG1 must be stretched to G ( where then G is not on line OB ) so that GG1 = OA .

For both motions there is one position where point G is on line OB and is not

needed GG1 to be stretched . This position is the common point P of the two circles which is their point of intersection . At this point P exists only rotation and is not needed GG1 to be stretched . This means that point P lies on the circle ( G , GG1 = OA ) , or GP = OA . Point A of angle < BOA is verged through two different and opposite motions , i.e.

1 . From point A to point Ao where is done a parallel translation of CA to the new position EAo , this is for all angles equal to 90 , and from this position

to the new position EA by rotating EAo to the new position EA having always

the distance E1 E2 = OA .

This motion is taking place on a circle of centre E1 and radius E1 E2 . 2. From point F , where OF = 2. OA , is done a parallel translation of AF to FAo, and from this position to the new position FA by rotating FAo to FA having always the distance F1 F2 = OA . The t

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